Clck to Vew Mathcad Document 2011 Knovel Corp. Buldng Structural Desgn. homas P. Magner, P.E. 2011 Parametrc echnology Corp. Chapter 3: Renforced Concrete Slabs and Beams 3.2 Renforced Concrete Beams - Sze Selecton Dsclamer Whle Knovel and Parametrc echnology Corporaton (PC) have made every effort to ensure that the calculatons, engneerng solutons, dagrams and other nformaton (collectvely Soluton ) presented n ths Mathcad worksheet are sound from the engneerng standpont and accurately represent the content of the book on whch the Soluton s based, Knovel and PC do not gve any warrantes or representatons, express or mpled, ncludng wth respect to ftness, ntended purpose, use or merchantablty and/or correctness or accuracy of ths Soluton. Array orgn: ORIGIN 1 Descrpton hs applcaton determnes the szes of rectangular beams to satsfy the flexural requrements, shear requrements and mnmum thckness lmts of ACI 318-89 usng the strength desgn method of ACI 318-89. he requred nput ncludes the strength of the concrete and the renforcement, the maxmum values for factored load bendng moments and shears, and the span lengths and types of span for the spans whch wll determne mnmum depths. hree beam szes are shown for llustratve purposes, however any practcal number of spans may be entered at one tme. Reference: ACI 318-89 "Buldng Code Requrements for Renforced Concrete." (Revsed 1992) Input Input Varables Enter Mu, Vu, L and Spanype as vectors wth the number of rows equal to the number of beam szes to be determned. Page 1 of 14
Crtcal factored moments: M u [ 190 85 75] kp ft Crtcal factored shears: V u [ 13 6.5 10.5 ] kp Span length, clear span for monolthc constructon: L [ 20 20 22 ] ft Span type: Spanype [ 1 2 1] he moment, shear, span length and span type are the crtcal values that determne the beam sze. he crtcal moment and shear do not necessarly occur at the same locaton. Enter values for bmn, hmax, R, and d'. Specfed mnmum member wdth: b mn 8n Page 2 of 14
Specfed maxmum permssble member thckness: Specfed maxmum rato of h/b: Estmated dstance from the centrod of the tenson renforcement to the extreme fber n tenson: h max 30 n R 2 d' 2.5 n Computed Varables b h d Mn Vn wdth of compresson face of member overall thckness of member dstance from extreme compresson fber to centrod of tenson renforcement useable moment capacty at factored load useable shear capacty at factored load Page 3 of 14
Materal Propertes and Constants Enter values for f'c, fy, wc, kv and ks f dfferent from that shown. Specfed compressve strength of concrete: Specfed yeld strength of renforcement (fy may not exceed 60 ks, ACI 318 11.5.2): f' c 4ks 60 ks Unt weght of concrete: Shear strength reducton factor (For lghtweght concrete kv = 1, for normal weght, kv = 0.75, for alllghtweght and sand-lghtweght concrete, kv = 0.85 (ACI 318, 11.2.1.2.)): w c 145 pcf k v 1 Enter factor for computng shear strength of strrups. (For 50% of maxmum shear renforcement stress and mnmum spacng at d/2, ks = 1. For 100% of maxmum shear renforcement stress and mnmum strrup spacng at d/4, ks = 2. (ACI 318, 11.5.4, 11.5.4.3)): Modulus of elastcty of renforcement (ACI 318, 8.5.2): Stran n concrete at compresson falure (ACI 318, 10.3.2): Strength reducton factor for flexure (ACI 318, 9.3.2.1): k s 1 E s 29000 ks ε c 0.003 ϕ f 0.9 Page 4 of 14
Strength reducton factor for shear (ACI 318, 9.3.2.3): Szng factor for roundng dmensons (to a multple of SzF): ϕ v 0.85 SzF 2 n Lmt the value of f'c for computng shear and development lengths to 10 ks by substtutng f'c_max for f'c n formulas for computng shear (ACI 318, 11.1.2, 12.1.2): f' c_max f f' c > 10 ks, 10 ks, f' c he followng values are computed from the entered materal propertes. Nomnal "one way" shear strength per unt area n concrete (ACI 318, 11.3.1.1, Eq (11-3), 11.5.4.3): v c f' c_max k v 2 ps v c = 126 ps ps Nomnal "one way" shear strength per unt area n concrete and shear renforcement (ACI 318, 11.3.1.1, Eq. (11-3), 11.5.4.3): v c f' c_max k v 2 ps v c = 126 ps ps v s f' c_max k s 4 ps v s = 253 ps ps Page 5 of 14
Modulus of elastcty of concrete for values of wc between 90 pcf and 155 pcf (ACI 318, 8.5.1): E c w 1.5 c 33 f' c ps E c = 3644 ks pcf ps Stran n renforcement at yeld stress: ε y ε y = 0.00207 E s Factor used to calculate depth of equvalent rectangular stress block (ACI 318, 10.2.7.3): β 1 f f' f' c 4 ks c 4 ks f' c 8 ks, 0.85 0.05, f f' c 4 ks, 0.85, 0.65 ks β 1 = 0.850 Renforcement rato producng balanced stran condtons (ACI 318, 10.3.2): ρ b β 1 0.85 f' c E s ε c ρ b = 2.851% E s ε c + Maxmum renforcement rato (ACI 318, 10.3.3): ρ max 3 4 ρ b ρ max = 2.138% Page 6 of 14
Preferred renforcement rato: ρ pref 0.5 ρ max ρ pref = 1.069% Mnmum renforcement rato for beams (ACI 318, 10.5.1, Eq. (10-3)): ρ mn 200 lbf ρ = n 2 mn 0.333% Shrnkage and temperature renforcement rato (ACI 318, 7.12.2.1): ρ temp f 50 ks,.002, f 60 ks,.002.0002, f.0018 60 ks.0018 60 ks.0014,,.0014 60 ks ρ temp = 0.180% Flexural coeffcent K, for rectangular beams or slabs, as a functon of (ACI 318, 10.2): (Moment capacty Mn = K( F, where F = bd2) K (ρ) ϕ f ρ ρ 1 2 0.85 f' c Factors for adjustng mnmum beam and slab thckness hmn for use of lghtweght concrete and yeld strengths other than 60 ks (ACI 318, 9.5.2.1, see footnotes to able 9.5 (a)): Adjustment factor for mnmum thckness for concrete weghts between 90 and 120 pcf: q 1 f w c 112 pcf, 1.65 0.005, pcf f w c 120 pcf, 1.09, 1 w c Page 7 of 14
q 1 = 1 Adjustment factor for mnmum thckness for yeld strengths other than 60 ks: q 2 0.4 + 100 ks q 2 = 1 Adjustment factor for mnmum thckness combnng factors for concrete weght and for yeld strengths other than 60 ks: Q Soluton q 1 q 2 Q= 1 Mnmum requred shear area ShA (ACI 318, 9.3.2.3, 11.3.1.1, Eq. (11-3), 11.5.4.3): V u ORIGIN last M u ShA ϕ v v c + v s ShA = [ 40.304 20.152 32.553 ] n 2 Mnmum member thckness hmn (unless deflectons are checked) (ACI 318, 9.5.2.1): S Spanype k f S =0, 16, f S =1, 18.5, f S =2, 21, 8 Page 8 of 14
h mn Q L k h mn = [ 12.973 11.429 14.270 ] n Round hmn up to nearest upper multple of SzF unless lower multple s wthn 1/2%: h Rmn 0.995 h mn SzF cel SzF h Rmn = [ 14 12 16 ] n Requred secton coeffcent F (F = bd2): F M u = K ρ pref F [ 4361.024 1950.984 1721.457 ] n 3 Calculate requred member sze: Guess value of h: h mn + h max h 2 h = [ 21.486 20.714 22.135 ] n Page 9 of 14
hckness h requred to satsfy flexural requrement wth preferred rato of h/b: f1 ( h, F) root h, R ( h 2.5 n) 2 F h h f f1 h, F h f = [ 22.284 17.452 16.811 ] n Round h up or down to the nearest multple of SzF: h f_rd h f SzF floor + SzF 0.5 h f_rd = [ 22 18 16 ] n Member thckness h determned n step 3 or as lmted by hrmn or hmax: h f h f_rd h max, h max, f h f_rd h Rmn, h Rmn, h f_rd h = [ 22.000 18.000 16.000 ] n Member wdths determned by flexure and shear: b f F 2 2.5 n h b f = [ 11.469 8.121 9.446 ] n Page 10 of 14
b v ShA 2.5 n h b v = [ 2.067 1.300 2.411 ] n he larger member wdth determned by rato R or bmn: b 1 f b mn h h,, R b mn R b 1 = [ 11 9 8 ] n he largest member wdth determned by shear, flexure, rato R or bmn: b 2 f b v b f, f b v b 1, b v, b 1, f b f b 1, b f, b 1 b 2 = [ 11.469 9.000 9.446 ] n Requred member wdth b rounded up to the nearest multple of SzF, unless lower multple s wthn 1/2%: b 0.995 b 2 SzF cel SzF b = [ 12 10 10] n Effectve depth to the centrod of the tenson renforcement: d h d' d = [ 19.5 15.5 13.5] n Page 11 of 14
heoretcal renforcement rato requred for flexure: ρ 1 1 2 M u 1 0.85 f' c ϕ f b d 2 0.85 f' c = ρ 1 [ 1.016 % 0.850 % 1.003% ] he larger of the theoretcal renforcement rato or the mnmum renforcement rato: ρ f ρ 1 3,, 4 ρ mn 4 3 ρ 1 f ρ 1 ρ,, mn ρ 1 ρ mn = ρ [ 1.016 % 0.850 % 1.003% ] Renforcement areas: A s ρ b d A s = [ 2.379 1.317 1.354 ] n 2 Useable moment capacty at factored load: ϕm n 2 K ρ b d ϕm n = [ 190.000 85.000 75.000 ] kp ft Page 12 of 14
Useable shear capacty at factored load: ϕv n ϕ v v c + v s b d ϕv n = [ 75.477 49.996 43.545 ] kp Useable shear capacty of concrete: ϕv c ϕ v v c b d ϕv c = [ 25.159 16.665 14.515 ] kp Summary Computed Varables Useable moment capacty at factored load: ϕm n = [ 190 85 75 ] kp ft Useable shear capacty at factored load: ϕv n = [ 75.477 49.996 43.545 ] kp Useable shear capacty of concrete: ϕv c = [ 25.159 16.665 14.515 ] kp Beam dmensons selected, renforcement ratos, renforcement areas, and mnmum and maxmum permssble values Member wdth: b = [ 12 10 10] n Member thckness: h = [ 22 18 16] n Page 13 of 14
Renforcement areas: A s = [ 2.379 1.317 1.354 ] n 2 Renforcement rato: ρ = [ 1.016 % 0.850 % 1.003% ] Mnmum requred renforcement rato: ρ mn = 0.333% Maxmum permssble renforcement rato: ρ max = 2.138% User Notces Equatons and numerc solutons presented n ths Mathcad worksheet are applcable to the specfc example, boundary condton or case presented n the book. Although a reasonable effort was made to generalze these equatons, changng varables such as loads, geometres and spans, materals and other nput parameters beyond the ntended range may make some equatons no longer applcable. Modfy the equatons as approprate our parameters fall outsde of the ntended range. For ths Mathcad worksheet, the global varable defnng the begnnng ndex dentfer for vectors and arrays, ORIGIN, s set as specfed n the begnnng of the worksheet, to ether 1 or 0. If ORIGIN s set to 1 and you copy any of the formulae from ths worksheet nto your own, you need to ensure that your worksheet s usng the same ORIGIN. Page 14 of 14